1008 Elevator （20 分）

最新推荐文章于 2022-02-10 21:00:30 发布

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最新推荐文章于 2022-02-10 21:00:30 发布

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分类专栏： PAT甲级试题

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本文探讨了一个具体的电梯调度问题，即在一栋高楼中，只有一部电梯如何高效地响应多个楼层的请求。通过分析电梯上行和下行的时间成本，提供了一种计算总时间开销的方法，适用于不超过100层的建筑。

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The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.

For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.

Input Specification:

Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.

Output Specification:

For each test case, print the total time on a single line.

Sample Input:

3 2 3 1

Sample Output:

#include <iostream>
using namespace std;
/*
 * 该题只需要注意电梯处在最底层不动的情况；
 * 上升时，上升楼层与所在楼层之差*6+5；
 * 下降时，下降楼层与所在楼层之差*4+5；
 */
int main() {
	int n,m,sum;
	int a[100001];
	cin>>n;
	for(int i=0;i<n;i++)
		cin>>a[i];
	m=sum=0;//定义初始楼层与时间都为0
	for(int i=0;i<n;i++){
		if(m<a[i])
			sum+=(a[i]-m)*6+5;
		else if(m>a[i])
			sum += (m - a[i]) * 4 + 5;
		else
			sum+=5;
		m=a[i];//记录变化后所在楼层
	}
	cout<<sum<<endl;
	return 0;
}